A circular platform is mounted on a frictionless vertical axle. Its radius \[\text{R}=\text{2m}\] and its moment of inertia about the axle is \[\text{200 kg }{{\text{m}}^{\text{2}}}\]. It is initially at rest. A \[\text{50 kg}\] man stands on the edge of the platform and begins to walk along the edge at the speed of \[\text{1 m/s}\] relative to the ground. Time taken by the man to complete one revolution is
A. $\dfrac{3\pi }{2}\text{s}$
B. $2\pi s$
C. $\dfrac{\pi }{2}\text{s}$
D. $\pi s$

Hint:In the given question as the circular platform is at rest hence if the man and platform are considered as systems then we can say there is none of the force applied on the disc which can generate torque.

Complete answer:
Given, the platform is at rest initially.
Radius of platform \[\text{R}=\text{2m}\]
Moment of inertia of platform \[\text{I}=\text{200 kg }{{\text{m}}^{\text{2}}}\]
Mass of man \[\text{m}=\text{50 kg}\]

Walking speed of man along the edge of the platform \[\text{v}=\text{1m/s}\] relative to ground.

If man and platform are considered as systems, then outside the system there is none of the force applied on disc which can generate torque.

So external torque \[\tau {}_{\text{external}}=0\]

Thus the angular momentum of man and platform will remain conserved.

By using principle of conservation of angular momentum

Initial angular momentum \[\left( {{L}_{I}} \right)=\] Final angular momentum \[\left( {{L}_{f}} \right)\]
\[\Rightarrow mvR=I\omega \]
\[\Rightarrow 50\times 1\times 2=200\times \omega \]
\[\Rightarrow \text{ }\!\!\omega\!\!\text{ }=\dfrac{1}{2}\text{rad/s}\](In the opposite direction of man walks)
So the $A=\pi {{r}^{2}}$ velocity of platform \[{{V}_{p}}=\] angular velocity of platform \[\times R\]
\[\Rightarrow {{\text{V}}_{\text{p}}}\text{=}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ 2=1m/s}\]

So the velocity of man with respect to the platform \[{{V}_{m/p}}=\text{ }{{V}_{m/g}}~-\text{ }{{V}_{m/p}}\]
\[\Rightarrow {{\text{V}}_{\text{m/p}}}\text{=1-(-1)=2m/s}\]

If T is time taken by man to complete one revolution, then

Distance cover by man in revolution \[=2\pi r=4\pi \]
Time taken to complete one revolution $\left( \text{T} \right)\text{=}\dfrac{\text{distance}}{\text{velocity}}\text{=}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{2}}\text{=2 }\!\!\pi\!\!\text{ }\!\!~\!\!\text{ seconds}$

Therefore, the correct choice is: (B) $2\pi s$.

Note:
As the platform is at rest so we have to calculate the angular velocity of platform with direction and then velocity of platform with direction (direction gives us sign ) which gives the velocity of man with respect to platform.